20105ee131A_1_ps6_2010fall_ZZsol

20105ee131A_1_ps6_2010fall_ZZsol - EE 131A Probability...

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EE 131A Problem Set 6 Probability Wednesday, November 10, 2010 Instructor: Lara Dolecek Due: Wednesday, November 17, 2010 Reading: Chapter 5 of Probability, Statistics, and Random Processes by A. Leon-Garcia 100 points total 1. Coin tossing and two RVs. Problem 5.2, page 288 of ALG Solution : (a) The underlying space S = { HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT } . Let W and Z be the number of heads obtained by Carlos and Michael respectively when each flip a fair coin twice, then the range of the pair ( W, Z ) is given by: S WZ = { (0 , 0) , (0 , 1) , (0 , 2) , (1 , 0) , (1 , 1) , (1 , 2) , (2 , 0) , (2 , 1) , (2 , 2) } The mapping between ( w, z ) and ( x, y ) can be done as x = w - z and y = w + z , therefore the range of the pair ( X, Y ) is given by: S XY = { (0 , 0) , ( - 1 , 1) , ( - 2 , 2) , (1 , 1) , (0 , 2) , ( - 1 , 3) , (2 , 2) , (1 , 3) , (0 , 4) } (b) The probabilities for all values of ( X, Y ) are given by: P [ X = 0 , Y = 0] = P [ W = 0 , Z = 0] = P [ W = 0] P [ Z = 0] = 1 / 16 P [ X = - 1 , Y = 1] = P [ W = 0 , Z = 1] = P [ W = 0] P [ Z = 1] = 1 / 8 P [ X = - 2 , Y = 2] = P [ W = 0 , Z = 2] = P [ W = 0] P [ Z = 2] = 1 / 16 P [ X = 1 , Y = 1] = P [ W = 1 , Z = 0] = P [ W = 1] P [ Z = 0] = 1 / 8 P [ X = 0 , Y = 2] = P [ W = 1 , Z = 1] = P [ W = 1] P [ Z = 1] = 1 / 4 P [ X = - 1 , Y = 3] = P [ W = 1 , Z = 2] = P [ W = 1] P [ Z = 2] = 1 / 8 P [ X = 2 , Y = 2] = P [ W = 2 , Z = 0] = P [ W = 2] P [ Z = 0] = 1 / 16 P [ X = 1 , Y = 3] = P [ W = 2 , Z = 1] = P [ W = 2] P [ Z = 1] = 1 / 8 P [ X = 0 , Y = 4] = P [ W = 2 , Z = 2] = P [ W = 2] P [ Z = 2] = 1 / 16 (c) Based on the above probabilities, we have: P [ X + Y = 1] = 0 P [ X + Y = 2] = P [ X = 1 , Y = 1] + P [ X = 0 , Y = 2] + P [ X = - 1 , Y = 3] = 1 / 2 2. Coin tossing and two RVs and pdfs. Problem 5.10, page 289 of ALG Solution : (a) When using a fair coin, p X,Y ( x, y ) is sketched in Figure 1: (b)Based on p X,Y ( x, y ), the marginal pmf p X ( x ) and p Y ( y ) can be found as follows: p X ( x ) = 1 / 16 x = - 2 1 / 4 x = - 1 3 / 8 x = 0 1 / 4 x = 1 1 / 16 x = 2 1
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-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 0 0.5 1 1.5 2 2.5 3 3.5
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